Apply linear & quadratic equations

1. Revision of solving equations

2. Revision of simultaneous equations


ID is: 3306 Seed is: 3693

Points of intersection

Graph the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x24y=x+2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x24 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x24.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is −4. The coordinates of this point are (0;4).

The x-intercepts are the roots of the equation x24=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 0. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. The two numbers are 2 and 2. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x240=x2+2x2x4

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+2)2(x+2)0=(x2)(x+2)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=2 and x=2. The coordinates of these roots are (2;0) and (2;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=x+2 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 2. Its coordinates are: (0;2).

We will work out the root of the equation 0=x+2.

0=x+2=0x=2x=2

The root is: x=2. It's coordinates are: (2;0).

We will now plot the equation y=x+2 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (2;0) and (3;5).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x24 and y=x+2. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (2;0) and (3;5).


Submit your answer as: and

ID is: 3306 Seed is: 4389

Points of intersection

Sketch the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x2+2x3y=x+1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x2+2x3 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x2+2x3.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is −3. The coordinates of this point are (0;3).

The x-intercepts are the roots of the equation x2+2x3=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 2. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 3 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+2x30=x2+3xx3

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x+3)(x+3)0=(x1)(x+3)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=1 and x=3. The coordinates of these roots are (1;0) and (3;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=x+1 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 1. Its coordinates are: (0;1).

We will work out the root of the equation 0=x+1.

0=x+1=0x=1x=1

The root is: x=1. It's coordinates are: (1;0).

We will now plot the equation y=x+1 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (1;0) and (4;5).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x2+2x3 and y=x+1. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (1;0) and (4;5).


Submit your answer as: and

ID is: 3306 Seed is: 7207

Points of intersection

Sketch the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

y=x27x+6y=2x8
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The points of intersection are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.


STEP: Plot the graph of y=x27x+6 on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: y=x27x+6.

To plot this graph, we need to find the coordinates of the points where the graph will cross the x- and y-axes. These points are the x- and y-intercepts, respectively. We will then use these points to plot the graph.

The y-intercept is the quantity in the equation which is not multiplied by x. For this equation, the y-intercept is 6. The coordinates of this point are (0;6).

The x-intercepts are the roots of the equation x27x+6=0. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 7. This is the coefficient of the middle term. The product of the two numbers must be 6. This is the product of the coefficient of the first term 1 and the value of the last term 6. The two numbers are 1 and 6. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x27x+60=x2x6x+6

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

0=x(x1)6(x1)0=(x6)(x1)

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are x=6 and x=1. The coordinates of these roots are (6;0) and (1;0).

Using these x- and y-intercepts we will draw the graph:


STEP: Plot the graph of y=2x8 on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the y-intercept. After that we will calculate root for the linear equation. We remember that the y-intercept is the quantity in the equation which is not multiplied by x.

The y-intercept is 8. Its coordinates are: (0;8).

We will work out the root of the equation 0=2x8.

0=2x8=02x=8x=4

The root is: x=4. It's coordinates are: (4;0).

We will now plot the equation y=2x8 on the same Cartesian plane we have used for the graph of the quadratic equation.


STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: (7;6) and (2;4).

We have used a graphical method to determine the solutions of the simultaneous equations: y=x27x+6 and y=2x8. The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are (7;6) and (2;4).


Submit your answer as: and

ID is: 1401 Seed is: 791

Simultaneous equations: solving by substitution

Solve simultaneously for x and y. You should get two coordinate pairs, (x;y), for your answers.

2x=y16x2+y=x12
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate y.

2x=y16y=2x16

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (2x16), into the quadratic equation in place of y. Be sure to use brackets! Then simplify the equation to get standard form.

x2+y=x12x2+(2x16)=x12x2+2x16=x12x2+3x4=0to keep the quadratic term positivemove the terms onto the left side

STEP: Solve for the values of x
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two x-values.

x2+3x4=0(x1)(x+4)=0x=1 and x=4

STEP: Find the value of y for each of the x-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of y for these equations. We must substitute in the values x=1 and x=4. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=1:y=2(1)16y=14If x=4:y=2(4)16y=24

Finally we have the answers: if x=1 then y=14 and if x=4 then y=24.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (1;14) and (4;24).


Submit your answer as: and

ID is: 1401 Seed is: 2627

Simultaneous equations: solving by substitution

Solve simultaneously for x and y. You should get two coordinate pairs, (x;y), for your answers.

x1=3y+5y22=x+3y
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

x1=3y+5x=3y+6

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (3y+6), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

y22=x+3yy22=(3y+6)+3yy22=60=y24to keep the quadratic term positivemove the terms onto the right side

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

0=y240=(y2)(y+2)y=2 and y=2

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=2 and y=2. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=2:x=3(2)+6x=0If y=2:x=3(2)+6x=12

Finally we have the answers: if x=0 then y=2 and if x=12 then y=2.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (0;2) and (12;2).


Submit your answer as: and

ID is: 1401 Seed is: 1135

Simultaneous equations: solving by substitution

Solve simultaneously for x and y. You should get two coordinate pairs, (x;y), for your answers.

x+6y+3=0x+y2=2y3
INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.
Answer: The solutions are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.


STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate x.

x+6y+3=0x=6y3

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, (6y3), into the quadratic equation in place of x. Be sure to use brackets! Then simplify the equation to get standard form.

x+y2=2y3(6y3)+y2=2y3y26y3=2y3y24y=0to keep the quadratic term positivemove the terms onto the left side

STEP: Solve for the values of y
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two y-values.

y24y=0y(y4)=0y=4 and y=0

STEP: Find the value of x for each of the y-values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of x for these equations. We must substitute in the values y=4 and y=0. We can use either equation to do this, so we will choose the easier option: the linear equation.

If y=4:x=6(4)3x=27If y=0:x=6(0)3x=3

Finally we have the answers: if x=27 then y=4 and if x=3 then y=0.

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: (27;4) and (3;0).


Submit your answer as: and

ID is: 3307 Seed is: 1343

Simultaneous equations

Solve the following equations simultaneously:

y=x27x+9y=2x+5
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x+5 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x27x+9linear equationy=2x+5

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

2x+5=x27x+90=x27x+9+2x50=x27x+2x+950=x25x+4

STEP: Solve the equation x25x+4=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. The two numbers are 1 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x25x+40=x2x4x+4

We now group the first two terms together and group the last terms together and factorise.

0=x2x4x+40=x(x1)4(x1)

The last thing now is to factor the highest common factor.

0=(x4)(x1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 4 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=4:

y=2x+5=3

For x=1:

y=2x+5=3

The solutions are x=4 together with y=3 , and x=1 with y=3. In coordinates pairs, the solutions are (4;3) and (1;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;3) and (1;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;3) and (1;3).


Submit your answer as: and

ID is: 3307 Seed is: 6933

Simultaneous equations

Solve the following equations simultaneously:

y=x2+4x+3y=x1
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+4x+3linear equationy=x1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

x1=x2+4x+30=x2+4x+3+x+10=x2+4x+x+3+10=x2+5x+4

STEP: Solve the equation x2+5x+4=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. The two numbers are 1 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+5x+40=x2+x+4x+4

We now group the first two terms together and group the last terms together and factorise.

0=x2+x+4x+40=x(x+1)+4(x+1)

The last thing now is to factor the highest common factor.

0=(x+4)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 1 and 4.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=1:

y=x1=0

For x=4:

y=x1=3

The solutions are x=1 together with y=0 , and x=4 with y=3. In coordinates pairs, the solutions are (1;0) and (4;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;0) and (4;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;0) and (4;3).


Submit your answer as: and

ID is: 3307 Seed is: 9960

Simultaneous equations

Solve for x and y:

y=x2+5x+7y=x1
INSTRUCTION: You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x1 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+5x+7linear equationy=x1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

x1=x2+5x+70=x2+5x+7+x+10=x2+5x+x+7+10=x2+6x+8

STEP: Solve the equation x2+6x+8=0
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 8. This is the product of the coefficient of the first term 1 and the value of the last term 8. The two numbers are 2 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2+6x+80=x2+2x+4x+8

We now group the first two terms together and group the last terms together and factorise.

0=x2+2x+4x+80=x(x+2)+4(x+2)

The last thing now is to factor the highest common factor.

0=(x+4)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 2 and 4.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

For x=2:

y=x1=1

For x=4:

y=x1=3

The solutions are x=2 together with y=1 , and x=4 with y=3. In coordinates pairs, the solutions are (2;1) and (4;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (2;1) and (4;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (2;1) and (4;3).


Submit your answer as: and

ID is: 3303 Seed is: 3168

Simultaneous equations

Determine the solution of the following equations:

y=3x22x1y=x+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x22x1linear equationy=x+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x+3=3x22x10=3x22x1+x30=3x22x+x130=3x2x4

STEP: Solve the equation 0=3x2x4
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 12. This is the product of the coefficient of the first term 3 and the value of the last term 4. The two numbers are 3 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x2x40=3x2+3x4x4

We now group the first two terms together and group the last terms together and factorise.

0=3x2+3x4x40=3x(x+1)4(x+1)

The last thing now is to factor the highest common factor.

0=(3x4)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 43 and 1.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=43:

y=x+3=53

For x=1:

y=x+3=4

The solutions are x=43 together with y=53 , and x=1 with y=4. In coordinate pairs, the solutions are (43;53) and (1;4).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (43;53) and (1;4). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (43;53) and (1;4).


Submit your answer as: and

ID is: 3303 Seed is: 5332

Simultaneous equations

Solve for x and y:

y=2x2+10x+9y=x3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=2x2+10x+9linear equationy=x3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x3=2x2+10x+90=2x2+10x+9+x+30=2x2+10x+x+9+30=2x2+11x+12

STEP: Solve the equation 0=2x2+11x+12
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 11. This is the coefficient of the middle term. The product of the two numbers must be 24. This is the product of the coefficient of the first term 2 and the value of the last term 12. The two numbers are 3 and 8. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=2x2+11x+120=2x2+3x+8x+12

We now group the first two terms together and group the last terms together and factorise.

0=2x2+3x+8x+120=x(2x+3)+4(2x+3)

The last thing now is to factor the highest common factor.

0=(x+4)(2x+3)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 32 and 4.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=32:

y=x3=32

For x=4:

y=x3=1

The solutions are x=32 together with y=32 , and x=4 with y=1. In coordinate pairs, the solutions are (32;32) and (4;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (32;32) and (4;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (32;32) and (4;1).


Submit your answer as: and

ID is: 3303 Seed is: 745

Simultaneous equations

Solve the following equations simultaneously:

y=3x2+7x+1y=2x+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting 2x+3 in place of y in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=3x2+7x+1linear equationy=2x+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x+3=3x2+7x+10=3x2+7x+12x30=3x2+7x2x+130=3x2+5x2

STEP: Solve the equation 0=3x2+5x2
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 6. This is the product of the coefficient of the first term 3 and the value of the last term 2. The two numbers are 6 and 1. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=3x2+5x20=3x2+6xx2

We now group the first two terms together and group the last terms together and factorise.

0=3x2+6xx20=3x(x+2)(x+2)

The last thing now is to factor the highest common factor.

0=(3x1)(x+2)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 13 and 2.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

We will substitute each x-value into the linear equation. This will give us the corresponding values of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.

For x=13:

y=2x+3=113

For x=2:

y=2x+3=1

The solutions are x=13 together with y=113 , and x=2 with y=1. In coordinate pairs, the solutions are (13;113) and (2;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (13;113) and (2;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (13;113) and (2;1).


Submit your answer as: and

ID is: 3308 Seed is: 1517

Simultaneous equations

Solve the following equations simultaneously:

y=x2+4x+2x=y2
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y2 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x2+4x+2linear equationx=y2

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y2)2+4(y2)+20=(y2)2+4(y2)+2y0=y24y+4+(4y8)+2y0=y24y+4+4y8+2y0=y24y+4yy+48+20=y2y+48+20=y2y4+20=y2y2

STEP: Solve the equation 0=y2y2
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 1. This is the coefficient of the middle term. The product of the two numbers must be 2. This is the product of the coefficient of the first term 1 and the value of the last term 2. The two numbers are 1 and 2. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2y20=y2+y2y2

We now group the first two terms together and group the last terms together and factorise.

0=y2+y2y20=y(y+1)2(y+1)

The last thing now is to factor the highest common factor.

0=(y2)(y+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 2 and -1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=2:

x=y2=0

For y=1:

x=y2=3

The solutions are x=0 together with y=2 , and x=3 with y=1. In coordinates pairs, the solutions are (0;2) and (3;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (0;2) and (3;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (0;2) and (3;1).


Submit your answer as: and

ID is: 3308 Seed is: 5444

Simultaneous equations

Determine the solution of the following equations:

y=x22x+1x=y1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y1 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x22x+1linear equationx=y1

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y1)22(y1)+10=(y1)22(y1)+1y0=y22y+1(2y2)+1y0=y22y+12y+2+1y0=y22y2yy+1+2+10=y24yy+1+2+10=y25y+1+2+10=y25y+3+10=y25y+4

STEP: Solve the equation 0=y25y+4
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 5. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 1 and the value of the last term 4. The two numbers are 1 and 4. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y25y+40=y2y4y+4

We now group the first two terms together and group the last terms together and factorise.

0=y2y4y+40=y(y1)4(y1)

The last thing now is to factor the highest common factor.

0=(y4)(y1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 4 and 1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=4:

x=y1=3

For y=1:

x=y1=0

The solutions are x=3 together with y=4 , and x=0 with y=1. In coordinates pairs, the solutions are (3;4) and (0;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;4) and (0;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (3;4) and (0;1).


Submit your answer as: and

ID is: 3308 Seed is: 5195

Simultaneous equations

Solve for x and y:

y=x25x+3x=y+5
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+5 in place of x in the quadratic equation and then simplify.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

quadratic equationy=x25x+3linear equationx=y+5

This means we want two pairs of x and y values which solve both equations.

We will substitute the value of x from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
y=(y+5)25(y+5)+30=(y+5)25(y+5)+3y0=y2+10y+25(5y+25)+3y0=y2+10y+255y25+3y0=y2+10y5yy+2525+30=y2+5yy+2525+30=y2+4y+2525+30=y2+4y+3

STEP: Solve the equation 0=y2+4y+3
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 3. This is the product of the coefficient of the first term 1 and the value of the last term 3. The two numbers are 1 and 3. We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2+4y+30=y2+y+3y+3

We now group the first two terms together and group the last terms together and factorise.

0=y2+y+3y+30=y(y+1)+3(y+1)

The last thing now is to factor the highest common factor.

0=(y+3)(y+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are -1 and -3.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

We will substitute each y-value into the linear equation. This will give us the corresponding values of x.

NOTE: We can do that by substituting y into any of the original equations. It is easier to use the linear one.

For y=1:

x=y+5=4

For y=3:

x=y+5=2

The solutions are x=4 together with y=1 , and x=2 with y=3. In coordinates pairs, the solutions are (4;1) and (2;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;1) and (2;3). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;1) and (2;3).


Submit your answer as: and

ID is: 1399 Seed is: 7896

Simultaneous equations: intersection points

The graph below shows the equations:

y=x2+2andy=x243x41

Find the intersection points of the curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity x2+2; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is x2+2and:yis also x243x41Therefore: x2+2=x243x41

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 4. So multiply both sides of the equation by 4 in order to cancel the denominators. Then arrange the equation in standard form.

(4)(x2+2)=(x243x41)(4)2x+8=x23x40=x2x12


STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2x120=(x4)(x+3)x=3 and x=4

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=3 and x=4 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=3:y=(3)2+2y=72If x=4:y=(4)2+2y=0

Finally! The solutions are (3;72) and (4;0).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (3;72) and (4;0).


Submit your answer as: and

ID is: 1399 Seed is: 4269

Simultaneous equations: intersection points

The graph below shows the equations:

y=x3+43andy=x23x43

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity x3+43; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is x3+43and:yis also x23x43Therefore: x3+43=x23x43

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(x3+43)=(x23x43)(3)x+4=x23x40=x22x8


STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x22x80=(x4)(x+2)x=2 and x=4

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=2 and x=4 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=2:y=(2)3+43y=2If x=4:y=(4)3+43y=0

Finally! The solutions are (2;2) and (4;0).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (2;2) and (4;0).


Submit your answer as: and

ID is: 1399 Seed is: 2543

Simultaneous equations: intersection points

The graph below shows the equations:

y=20x3+16andy=4x23+4x3+16

Determine the points of intersection for these two curves.

INSTRUCTION:
  • Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
  • It does not matter which coordinate pair you type first.
  • Do not round your answers.
Answer: The points of intersection are and .
coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.


STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation y is equal to the quantity 20x3+16; and at the same time, the quadratic equation also contains the value y. Substitute like this:

y is 20x3+16and:yis also 4x23+4x3+16Therefore: 20x3+16=4x23+4x3+16

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is 3. So multiply both sides of the equation by 3 in order to cancel the denominators. Then arrange the equation in standard form.

(3)(20x3+16)=(4x23+4x3+16)(3)20x+48=4x2+4x+480=4x216x

In this case, there is a common factor of −4 for all of the terms. Divide out this factor before factorising.

0=x2+4x

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two x-values.

0=x2+4x0=x(x+4)x=4 and x=0

STEP: Use the values of x to find the values of y
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of y for these equations. Substitute in the values x=4 and x=0 into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

If x=4:y=20(4)3+16y=323If x=0:y=20(0)3+16y=16

Finally! The solutions are (4;323) and (0;16).

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are (4;323) and (0;16).


Submit your answer as: and

ID is: 3305 Seed is: 6478

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x2x+6y=3x+9

Which of the following equations shows the correct substitution step?

A 3x+9=x2x+6
B 3x+9=x2x
C 3x9=x2x+6
D y=(3x+9)2(3x+9)+6
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute 3x+9 in place of y in the quadratic equation y=x2x+6.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x2x+6linear equationy=3x+9

We need to substitute 3x+9 from the linear equation in place of y in the quadratic equation.

3x+9=x2x+6
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is 3x+9=x2x+6 which is choice A.


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ID is: 3305 Seed is: 5680

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x2+2x5x=y+3

Which of the following equations shows the correct substitution step?

A y=y2+2y5
B y=(y+3)2+2(y+3)5
C 3=x2+2x5
D y=y+32+2y+35
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute y+3 in place of x in the quadratic equation y=x2+2x5.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x2+2x5linear equationx=y+3

We need to substitute y+3 from the linear equation in place of x in the quadratic equation.

y=(y+3)2+2(y+3)5
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for y. The question does not ask us to solve the simultaneous equations.

The correct substitution step is y=(y+3)2+2(y+3)5 which is choice B.


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ID is: 3305 Seed is: 8504

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

y=x24x1y=2x+2

Which of the following equations shows the correct substitution step?

A 2x+2=y24y1
B 2y+2=x24x1
C 1=x24x1
D 2x+2=x24x1
Answer: The correct substitution step is choice: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute 2x+2 in place of y in the quadratic equation y=x24x1.


STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

quadratic equationy=x24x1linear equationy=2x+2

We need to substitute 2x+2 from the linear equation in place of y in the quadratic equation.

2x+2=x24x1
NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for x. The question does not ask us to solve the simultaneous equations.

The correct substitution step is 2x+2=x24x1 which is choice D.


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ID is: 3289 Seed is: 7175

Equations with fractions

What are the values of x and y which solve these equations simultaneously?

2y+5x=3xyx,y04x=2y2

Your answer should be exact (do not round off).

Answer:

The solution is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is xy.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by xy, which is the LCD of the fractions. That will cancel all of the denominators.

2y+5x=3xyxy(2y+5x)=xy(3xy)5xxy+2yxy=32x+5y=3

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

2x+5y=34x=2y2

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by 2 so that the x-coefficients will be ready to cancel. Then we can eliminate them by adding the equations.

First modify things to set up the elimination:

2(2x+5y)=2(3)4x+10y=6

Now do the elimination and complete the solution to find y.

4x+10y=6second equationAdd the+(4x)=+(2y2)4x+4x+10y=2+6+2y10y=2y+48y=4y=12

Great - we have the first value, y=12.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of x. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

4x=2y24x=2(12)24x=1x=14

The answer is the pair of numbers x=14 and y=12. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force x,y0, which is why there are open intervals at the x- and y-intercepts of the fraction equation.

The values which solve these two equations are x=14 and y=12.


Submit your answer as: and

ID is: 3289 Seed is: 4822

Equations with fractions

Determine the values of a and b which solve these two equations:

11ab=3b+4aa,b02a2=4b

Your answer should be exact (do not round off).

Answer:

The solution is a= and b= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is ab.


STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by ab, which is the LCD of the fractions. That will cancel all of the denominators.

11ab=3b+4aab(11ab)=ab(3b+4a)11=4aab+3bab11=3a+4b

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

11=3a+4b2a2=4b

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations invite elimination, because the b-terms have equal and opposite coefficients. We can eliminate the b-terms by adding the equations.

Eliminate the b and solve for a:

11=3a+4bsecond equationAdd the+(2a2)=+(4b)2a2+11=3a4b+4b2a+9=3aa=9a=9

Great - we have the first value, a=9.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of b. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

2a2=4b2(9)2=4b4b=16b=4

The answer is the pair of numbers a=9 and b=4. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force a,b0, which is why there are open intervals at the a- and b-intercepts of the fraction equation.

The values which solve these two equations are a=9 and b=4.


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ID is: 3289 Seed is: 3358

Equations with fractions

What are the values of x and y which solve these equations simultaneously?

2x+2=2y3xy=3y4xx,y0

Your answer should be exact (do not round off).

Answer:

The solution is x= and y= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is xy.


STEP: Rearrange the second equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by xy, which is the LCD of the fractions. That will cancel all of the denominators.

3xy=3y4xxy(3xy)=xy(3y4x)3=4xxy+3yxy3=3x4y

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

2x+2=2y3=3x4y

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by 2 so that the y-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

2(2x+2)=2(2y)4x+4=4y

Now do the elimination and complete the solution to find x.

4x+4=4ysecond equationSubtract the(3)=(3x4y)3+44x=3x4y+4y4x+7=3xx=7x=7

Great - we have the first value, x=7.


STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of y. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

2x+2=2y2(7)+2=2y2y=12y=6

The answer is the pair of numbers x=7 and y=6. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force x,y0, which is why there are open intervals at the x- and y-intercepts of the fraction equation.

The values which solve these two equations are x=7 and y=6.


Submit your answer as: and

ID is: 3304 Seed is: 6455

Simultaneous equations: special outcomes

Solve for x and y:

y=3x2+5x+2y=x1
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting x1 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=3x2+5x+2 and the linear equation is y=x1. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
x1=3x2+5x+20=3x2+5x+2+x+10=3x2+5x+x+2+10=3x2+6x+3

STEP: Solve the equation 0=3x2+6x+3
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 9. This is the product of the coefficient of the first term 3 and the value of the last term 3. If we take 3 which is half of the value of 6 (the coefficient of the middle term) and square it, we get the product 9. In this case we will use only one number to factorise, that is, 3. We will re-write the middle term, forming two terms, using this value.

0=3x2+6x+30=3x2+3x+3x+3

We now group the first two terms together and group the last terms together and factorise.

0=3x2+3x+3x+30=3x(x+1)+3(x+1)

The last thing now is to factor the highest common factor.

0=(3x+3)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 1. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=x1=1(1)1=0

The solution is x=1 together with y=0. Using coordinate pairs, the solution is (1;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;0). This is the same solution we have calculated. This means we are 100% correct.

The solution is (1;0).


Submit your answer as:

ID is: 3304 Seed is: 9523

Simultaneous equations: special outcomes

Solve the following equations simultaneously:

y=2x2+x1y=3x3
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting 3x3 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=2x2+x1 and the linear equation is y=3x3. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
3x3=2x2+x10=2x2+x1+3x+30=2x2+x+3x1+30=2x2+4x+2

STEP: Solve the equation 0=2x2+4x+2
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 4. This is the coefficient of the middle term. The product of the two numbers must be 4. This is the product of the coefficient of the first term 2 and the value of the last term 2. If we take 2 which is half of the value of 4 (the coefficient of the middle term) and square it, we get the product 4. In this case we will use only one number to factorise, that is, 2. We will re-write the middle term, forming two terms, using this value.

0=2x2+4x+20=2x2+2x+2x+2

We now group the first two terms together and group the last terms together and factorise.

0=2x2+2x+2x+20=2x(x+1)+2(x+1)

The last thing now is to factor the highest common factor.

0=(2x+2)(x+1)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 1. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=3x3=3(1)3=0

The solution is x=1 together with y=0. Using coordinate pairs, the solution is (1;0).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;0). This is the same solution we have calculated. This means we are 100% correct.

The solution is (1;0).


Submit your answer as:

ID is: 3304 Seed is: 6181

Simultaneous equations: special outcomes

Solve for x and y in the following simultaneous equations:

y=x2+4x+6y=2x3
INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.
Answer: The solution is .
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting 2x3 in place of y in the quadratic equation and simplify. You must expect a special kind of solution.


STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is y=x2+4x+6 and the linear equation is y=2x3. This means we expect two pairs of x and y values which solve both equations.

We will substitute the value of y from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of y from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.
2x3=x2+4x+60=x2+4x+6+2x+30=x2+4x+2x+6+30=x2+6x+9

STEP: Solve the equation 0=x2+6x+9
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give 6. This is the coefficient of the middle term. The product of the two numbers must be 9. This is the product of the coefficient of the first term 1 and the value of the last term 9. If we take 3 which is half of the value of 6 (the coefficient of the middle term) and square it, we get the product 9. In this case we will use only one number to factorise, that is, 3. We will re-write the middle term, forming two terms, using this value.

0=x2+6x+90=x2+3x+3x+9

We now group the first two terms together and group the last terms together and factorise.

0=x2+3x+3x+90=x(x+3)+3(x+3)

The last thing now is to factor the highest common factor.

0=(x+3)(x+3)

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, 3. This root is the x-value of our solution. This x-value will have one corresponding y value.


STEP: Calculate the corresponding value of y
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of x, we will substitute it into the linear equation to calculate the corresponding value of y.

NOTE: We can do that by substituting x into any of the original equations. It is easier to use the linear one.
y=2x3=2(3)3=3

The solution is x=3 together with y=3. Using coordinate pairs, the solution is (3;3).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (3;3). This is the same solution we have calculated. This means we are 100% correct.

The solution is (3;3).


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ID is: 3852 Seed is: 9111

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=4yandx22xy=32
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=4y into x22xy=32 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=4yandx22xy=32

First, we can substitute 4y in place of x in the second equation:

(4y)22(4y)(y)=32

We can simplify this equation to solve for y:

(4y)22(4y)(y)=3216y28y2=328y2=32y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=4y.

For y=2:

x=4y=4(2)=8

And for y=2:

x=4y=4(2)=8

So the x-values are x=8 or x=8.

Therefore the correct coordinate pairs are (8;2) and (8;2).


Submit your answer as: and

ID is: 3852 Seed is: 4488

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx2+4xy=16
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x2+4xy=16 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx2+4xy=16

First, we can substitute 2y in place of x in the second equation:

(2y)2+4(2y)(y)=16

We can simplify this equation to solve for y:

(2y)2+4(2y)(y)=164y28y2=164y2=16y2=4

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=2 or y=2

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=2:

x=2y=2(2)=4

And for y=2:

x=2y=2(2)=4

So the x-values are x=4 or x=4.

Therefore the correct coordinate pairs are (4;2) and (4;2).


Submit your answer as: and

ID is: 3852 Seed is: 9619

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for x and y:

x=2yandx24xy=36
INSTRUCTIONS:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are
and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.


STEP: Substitute x=2y into x24xy=36 and solve for y
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for x and y:

x=2yandx24xy=36

First, we can substitute 2y in place of x in the second equation:

(2y)24(2y)(y)=36

We can simplify this equation to solve for y:

(2y)24(2y)(y)=364y28y2=364y2=36y2=9

y can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about y2. So we have to consider both the positive and the negative values as solutions to the equations.

y=3 or y=3

STEP: Use the y-values to find the x-values
[−1 point ⇒ 0 / 4 points left]

Each of the y-values that we have found will have a corresponding value of x, which can be found from the equation x=2y.

For y=3:

x=2y=2(3)=6

And for y=3:

x=2y=2(3)=6

So the x-values are x=6 or x=6.

Therefore the correct coordinate pairs are (6;3) and (6;3).


Submit your answer as: and

ID is: 3302 Seed is: 3615

Simultaneous equations with xy terms

Solve for x and y:

2y=5xxy+1x=y+3
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+3 in place of x in the hyperbolic equation and then solve for y.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=5xxy+1linear equationx=y+3

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of x from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 2y=5xxy+1 and substitute its value into x=y+3. We could still get the same answer, but the calculations might become too long.
2y=5(y+3)(y+3)y+10=5(y+3)(y+3)y+12y0=5(y+3)y(y+3)+12y0=y25y3y2y+15+10=y28y2y+15+10=y210y+15+10=y210y+16

STEP: Solve the equation 0=y210y+16
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y210y+160=(y2)(y8)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 2 and 8.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

By using the values of y, we will calculate the corresponding values of x. We will do so by substituting each y-value into the linear equation.

NOTE: We can still get the same values of x if we substitute the values y into the equation 2y=5xxy+1. If we do that, the calculations might become too long.

For y=2:

x=y+3=1

For y=8:

x=y+3=5

The solutions are x=1 together with y=2 , and x=5 with y=8. Using coordinate pairs, the solutions are (1;2) and (5;8).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (1;2) and (5;8). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (1;2) and (5;8).


Submit your answer as: and

ID is: 3302 Seed is: 5830

Simultaneous equations with xy terms

Solve for x and y in the following simultaneous equations:

2y=3xxy2y=x+4
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting x+4 in place of y in the hyperbolic equation and then solve for x.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=3xxy2linear equationy=x+4

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of y from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 2y=3xxy2 and substitute its value into y=x+4. We could still get the same answer, but the calculations might become too long.
2(x+4)=3xx(x+4)20=3xx(x+4)2+2(x+4)0=3x(x2+4x)2+(2x+8)0=x2x+2x2+80=x2+x2+80=x2+x+6make the first term positivedivide each term by -1 to 0=x2x6

STEP: Solve the equation 0=x2x6
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=x2x60=(x+2)(x3)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the x-values of our solutions. The x-values are 2 and 3.


STEP: Calculate the corresponding values of y
[−2 points ⇒ 0 / 6 points left]

By using the values of x, we will calculate the corresponding values of y. We will do so by substituting each x-value into the linear equation.

NOTE: We can still get the same values of y if we substitute the values x into the equation 2y=3xxy2. If we do that, the calculations might become too long.

For x=2:

y=x+4=2

For x=3:

y=x+4=7

The solutions are x=2 together with y=2 , and x=3 with y=7. Using coordinate pairs, the solutions are (2;2) and (3;7).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (2;2) and (3;7). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (2;2) and (3;7).


Submit your answer as: and

ID is: 3302 Seed is: 9258

Simultaneous equations with xy terms

Solve the following equations simultaneously:

2y=5xxy+2x=y+1
INSTRUCTION:
  • You should get two coordinate pairs in the form (x1;y1) and (x2;y2). You must enter each pair of coordinates in one of the input boxes provided.
  • It does not matter which coordinate pair you type first.
Answer:

The solutions are and .

coordinate
coordinate
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting y+1 in place of x in the hyperbolic equation and then solve for y.


STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

hyperbolic equation2y=5xxy+2linear equationx=y+1

This means we want two pairs of x and y-values which solve both equations.

We will substitute the value of x from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either x or y from 2y=5xxy+2 and substitute its value into x=y+1. We could still get the same answer, but the calculations might become too long.
2y=5(y+1)(y+1)y+20=5(y+1)(y+1)y+22y0=5(y+1)y(y+1)+22y0=y2+5yy2y5+20=y2+4y2y5+20=y2+2y5+20=y2+2y3

STEP: Solve the equation 0=y2+2y3
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.
0=y2+2y30=(y+3)(y1)

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the y-values of our solutions. The y-values are 3 and 1.


STEP: Calculate the corresponding values of x
[−2 points ⇒ 0 / 6 points left]

By using the values of y, we will calculate the corresponding values of x. We will do so by substituting each y-value into the linear equation.

NOTE: We can still get the same values of x if we substitute the values y into the equation 2y=5xxy+2. If we do that, the calculations might become too long.

For y=3:

x=y+1=4

For y=1:

x=y+1=0

The solutions are x=4 together with y=3 , and x=0 with y=1. Using coordinate pairs, the solutions are (4;3) and (0;1).

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points (4;3) and (0;1). These are the same solutions we have calculated. This means we are 100% correct.

The solutions are (4;3) and (0;1).


Submit your answer as: and

3. Practical applications


ID is: 3280 Seed is: 9570

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 90.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 90.
The numbers are consecutive numbers.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis90n1×n2=90

The correct equation for the first fact is n1×n2=90.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the second option.

The smaller numberis1 less than the larger numbern1=n21

This makes sense: consecutive numbers are separated by 1. The equation shows that with the 1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 90. n1×n2=90
The numbers are consecutive numbers. n1=n21

Submit your answer as: and

ID is: 3280 Seed is: 3556

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 56.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 56.
The numbers are consecutive numbers.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis56n1×n2=56

The correct equation for the first fact is n1×n2=56.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 56. n1×n2=56
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

ID is: 3280 Seed is: 8822

Setting up simultaneous equations

Here are facts about two numbers:

  • The product of the numbers is 132.
  • The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let n1 represent one of the numbers and n2 represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:
Fact about the numbers Equation
The product of the numbers is 132.
The numbers are consecutive numbers.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let n1 represent one of the numbers and n2 represent the other number". So we can break up the first fact like this:

The product of the numbersis132n1×n2=132

The correct equation for the first fact is n1×n2=132.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like 10 and 11, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is 1 more than the small number." Another is "the small number is 1 less than the larger number." Let's use the first option.

The larger numberis1 more than the smaller numbern2=n1+1

This makes sense: consecutive numbers are separated by 1. The equation shows that with the +1.

The correct answers are:

Fact about the numbers Equation
The product of the numbers is 132. n1×n2=132
The numbers are consecutive numbers. n2=n1+1

Submit your answer as: and

ID is: 381 Seed is: 6162

Word problems: distance, speed and time

Two missiles start out at two different launchers. The missiles are 1,306 km apart. The missiles start moving towards each other. One missile is moving at 289 km/h and the other missile at 364 km/h.

If both missiles started their journey at the same time, how long will they take to pass each other?

Answer: The two missiles pass each other after hours.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two missiles must be equal to the total distance when the missiles meet:

d1+d2=dtotald1+d2=1,306 km

STEP: Write equations to describe the motion of the missiles
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the missiles to meet. Let the time taken be t. Then we can write an expression for the distance each of the missiles travels:

For missile 1:

d1=s1tmissile is 289 km/hThe speed of the firstd1=289t

For missile 2:

d2=s2tmissile is 364 km/hThe speed of the secondd2=364t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=1,306(289t)+(364t)=1,306653t=1,306t=1,306653t=2

The missiles will meet after 2 hours.


Submit your answer as:

ID is: 381 Seed is: 1891

Word problems: distance, speed and time

Two boats start out at two different harbours. The boats are 1,085 km apart. The boats start moving towards each other. One boat is moving at 60 km/h and the other boat at 95 km/h.

If both boats started their journey at the same time, how long will they take to pass each other?

Answer: The two boats pass each other after hours.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two boats must be equal to the total distance when the boats meet:

d1+d2=dtotald1+d2=1,085 km

STEP: Write equations to describe the motion of the boats
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the boats to meet. Let the time taken be t. Then we can write an expression for the distance each of the boats travels:

For boat 1:

d1=s1tboat is 60 km/hThe speed of the firstd1=60t

For boat 2:

d2=s2tboat is 95 km/hThe speed of the secondd2=95t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=1,085(60t)+(95t)=1,085155t=1,085t=1,085155t=7

The boats will meet after 7 hours.


Submit your answer as:

ID is: 381 Seed is: 7539

Word problems: distance, speed and time

Two ships start out at two different ports. The ships are 676 km apart. The ships start moving towards each other. One ship is moving at 73 km/h and the other ship at 96 km/h.

If both ships started their journey at the same time, how long will they take to pass each other?

Answer: The two ships pass each other after hours.
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.


STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two ships must be equal to the total distance when the ships meet:

d1+d2=dtotald1+d2=676 km

STEP: Write equations to describe the motion of the ships
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

speed =distance time OR distance =speed ×time

We want to know the amount of time needed for the ships to meet. Let the time taken be t. Then we can write an expression for the distance each of the ships travels:

For ship 1:

d1=s1tship is 73 km/hThe speed of the firstd1=73t

For ship 2:

d2=s2tship is 96 km/hThe speed of the secondd2=96t

STEP: Solve the equations simultaneously for t
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of t.

d1+d2=676(73t)+(96t)=676169t=676t=676169t=4

The ships will meet after 4 hours.


Submit your answer as:

ID is: 392 Seed is: 3999

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 6 waffles and 4 hotdogs. Here are some facts about their lunch:

  • the total cost for the 6 waffles and 4 hotdogs is R270
  • a waffle costs N=5 more than a hotdog

What is the price for one waffle and the price for one hotdog?

Answer:

A waffle costs N= and a hotdog costs N= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a waffle and a hotdog). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

w=the price of a waffleh=the price of a hotdog

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 6 waffles and 4 hotdogs is N=270." We can use the expression 6w to represent the price of the 6 waffles. Similarly, the expression 4h represents the price of the 4 hotdogs. With these values we can write a full equation for the prices:

In words: the total cost for the 6 waffles and 4 hotdogs is N=270In maths: 6w+4h=270

Now we can use the second point. It says, "a waffle costs N=5 more than a hotdog." We can write this as an equation like this:

In words: a waffle costs N=5 more than a hotdogIn maths: w=h+5

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

6w+4h=270w=h+5

We could use elimination, but substitution is a better choice (because w is already the subject). Substitute the second equation into the first equation and solve!

6w+4h=2706(h+5)+4h=2706h+30+4h=27010h=27030h=24010=24

This means that the price of one hotdog is N=24.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for h to find the value of w (the price of a waffle). We can use either equation to do this, but the second one is easier to use (because w is already isolated).

w=h+5=24+5=29

The price for one waffle is N=29.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the waffle is N=29 while a hotdog costs N=24.


Submit your answer as: and

ID is: 392 Seed is: 8794

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 4 pizzas and 2 sandwiches. Here are some facts about their lunch:

  • the total cost for the 4 pizzas and 2 sandwiches is R176
  • a pizza costs N=8 more than a sandwich

What is the price for one pizza and the price for one sandwich?

Answer:

A pizza costs N= and a sandwich costs N= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a pizza and a sandwich). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

p=the price of a pizzas=the price of a sandwich

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 4 pizzas and 2 sandwiches is N=176." We can use the expression 4p to represent the price of the 4 pizzas. Similarly, the expression 2s represents the price of the 2 sandwiches. With these values we can write a full equation for the prices:

In words: the total cost for the 4 pizzas and 2 sandwiches is N=176In maths: 4p+2s=176

Now we can use the second point. It says, "a pizza costs N=8 more than a sandwich." We can write this as an equation like this:

In words: a pizza costs N=8 more than a sandwichIn maths: p=s+8

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

4p+2s=176p=s+8

We could use elimination, but substitution is a better choice (because p is already the subject). Substitute the second equation into the first equation and solve!

4p+2s=1764(s+8)+2s=1764s+32+2s=1766s=17632s=1446=24

This means that the price of one sandwich is N=24.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for s to find the value of p (the price of a pizza). We can use either equation to do this, but the second one is easier to use (because p is already isolated).

p=s+8=24+8=32

The price for one pizza is N=32.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the pizza is N=32 while a sandwich costs N=24.


Submit your answer as: and

ID is: 392 Seed is: 7693

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys 6 sandwiches and 4 hamburgers. Here are some facts about their lunch:

  • the total cost for the 6 sandwiches and 4 hamburgers is R330
  • a sandwich costs N=5 more than a hamburger

What is the price for one sandwich and the price for one hamburger?

Answer:

A sandwich costs N= and a hamburger costs N= .

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.


STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a sandwich and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

s=the price of a sandwichh=the price of a hamburger

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the 6 sandwiches and 4 hamburgers is N=330." We can use the expression 6s to represent the price of the 6 sandwiches. Similarly, the expression 4h represents the price of the 4 hamburgers. With these values we can write a full equation for the prices:

In words: the total cost for the 6 sandwiches and 4 hamburgers is N=330In maths: 6s+4h=330

Now we can use the second point. It says, "a sandwich costs N=5 more than a hamburger." We can write this as an equation like this:

In words: a sandwich costs N=5 more than a hamburgerIn maths: s=h+5

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

6s+4h=330s=h+5

We could use elimination, but substitution is a better choice (because s is already the subject). Substitute the second equation into the first equation and solve!

6s+4h=3306(h+5)+4h=3306h+30+4h=33010h=33030h=30010=30

This means that the price of one hamburger is N=30.


STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for h to find the value of s (the price of a sandwich). We can use either equation to do this, but the second one is easier to use (because s is already isolated).

s=h+5=30+5=35

The price for one sandwich is N=35.


STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the sandwich is N=35 while a hamburger costs N=30.


Submit your answer as: and

ID is: 3284 Seed is: 6079

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Gift buys some bananas and some sweets. The total cost for 2 bananas and 3 sweets is N=19.10. And each sweet costs N=0.30 less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be N=6.75. Or N=5. Or N=19.87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like N=4.30, or must a price always be a whole number like N=4.00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -N=5.50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3284 Seed is: 7695

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Bianca buys some mangoes and some packs of gum. The total cost for 2 mangoes and 6 packs of gum is N=29.40. And each pack of gum costs N=2.70 less than each mango. What is the price for one mango?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the price be a decimal number (or must it be an integer)?
  2. Can the price be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be N=6.75. Or N=5. Or N=19.87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like N=4.30, or must a price always be a whole number like N=4.00? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.


STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -N=5.50. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3284 Seed is: 8691

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are fruits and sweets for sale. Zenzile buys some mangoes and some packs of gum. The total number of items he buys is 8. Each mango costs N=3.80 and each pack of gum costs N=3.60. The total cost is N=29.20. How many mangoes did Zenzile buy?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of items be a decimal number (or must it be an integer)?
  2. Can the number of items be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of mangoes Zenzile buys. Can the number of items be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the number of items can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be N=6.75. Or N=5. Or N=19.87. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the number of items can be a decimal number or not. In other words, can the number of mangoes be a decimal number like 3.7 mangoes? Or must it be a whole number like 4 mangoes? This is different from the airtime example above: airtime can be a decimal number, but it is not possible to buy 3.5 mangoes! (You might think, "Wait! I can buy 3.5 mangoes, just cut one of them in half." But no one should expect to do this while shopping.)

The answer for the first question is: No, the number of items cannot be a decimal number.


STEP: Decide if the number of items can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the number of items can be negative, or must be postive. The number of mangoes cannot be negative. If Zenzile did not buy any mangoes at all, then the number of mangoes is zero. But is it impossible for him to buy 3 mangoes.

The answer to the second question is: No, the number of items cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3282 Seed is: 6884

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Thobeka is drawing a pattern of rectangles in her notebook. The number of rectangles in each figure is 2 more than the number of rectangles in the previous figure. If there are 7 rectangles in the third figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of rectangles be a decimal number (or must it be an integer)?
  2. Can the number of rectangles be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Thobeka's diagram. Can the number of rectangles be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176.3562 days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like 3.7, or must the number be a whole number like 4? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. 3.7 rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get 3.7 rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.


STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Thobeka drew any rectangles at all, then the answer to this question must be positive. There is no such thing as 4 rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3282 Seed is: 245

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Latifah is drawing a pattern of rectangles in her notebook. The number of rectangles in each figure is 4 more than the number of rectangles in the previous figure. If there are 16 rectangles in the third figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the number of rectangles be a decimal number (or must it be an integer)?
  2. Can the number of rectangles be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Latifah's diagram. Can the number of rectangles be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176.3562 days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like 3.7, or must the number be a whole number like 4? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. 3.7 rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get 3.7 rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.


STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Latifah drew any rectangles at all, then the answer to this question must be positive. There is no such thing as 4 rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

  1. No, it must be an integer
  2. No, it cannot be negative

Submit your answer as: and

ID is: 3282 Seed is: 7268

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Jacob is drawing a rectangle in his notebook. The length of the rectangle is 3 cm more than its width. The area of the rectangle is 28 cm2. What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:
  1. Can the width of the rectangle be a decimal number (or must it be an integer)?
  2. Can the width of the rectangle be a negative number?
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like 4.5? Can it be negative, like 5?


STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be 150 days of school, shame. Similarly, it is not possible to have 176.3562 days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like 3.5 cm, or must a distance be a whole number like 4 cm? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have 3.5 days of school, but a rectangle certainly can have a width of 3.5 cm. Or 5.927 cm. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.


STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room 5 metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

  1. Yes, it can be a decimal
  2. No, it cannot be negative

Submit your answer as: and

ID is: 382 Seed is: 7454

Word problems: test scores and simultaneous equations

Chinedu and Akinlabi are friends. Chinedu takes Akinlabi's history test paper and says: “I have 6 marks more than you do and the sum of both our marks is equal to 170. What are our marks?”

Answer:

Chinedu got marks and Akinlabi got marks for the history test paper.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

c= Chinedu's marka= Akinlabi's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than AkinlabiChinedu has 6 marksc=a+6(the total) is 170The sum of the marksc+a=170

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

c+a=170(a+6)+a=1702a=1706a=1642=82

This means that Akinlabi's mark is 82.


STEP: Find Chinedu's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Akinlabi's mark and one of the equations we have to find Chinedu's mark. The easiest way to do that is to substitute Akinlabi's mark back into the first equation:

c=a+6=(82)+6=88

The students achieved these marks: Chinedu earned 88 and Akinlabi earned 82.


Submit your answer as: and

ID is: 382 Seed is: 4953

Word problems: test scores and simultaneous equations

Abodunrin and Latifah are friends. Abodunrin takes Latifah's civil technology test paper and says: “I have 8 marks more than you do and the sum of both our marks is equal to 174. What are our marks?”

Answer:

Abodunrin got marks and Latifah got marks for the civil technology test paper.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

a= Abodunrin's markl= Latifah's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than LatifahAbodunrin has 8 marksa=l+8(the total) is 174The sum of the marksa+l=174

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

a+l=174(l+8)+l=1742l=1748l=1662=83

This means that Latifah's mark is 83.


STEP: Find Abodunrin's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Latifah's mark and one of the equations we have to find Abodunrin's mark. The easiest way to do that is to substitute Latifah's mark back into the first equation:

a=l+8=(83)+8=91

The students achieved these marks: Abodunrin earned 91 and Latifah earned 83.


Submit your answer as: and

ID is: 382 Seed is: 266

Word problems: test scores and simultaneous equations

Mkhuseli and Yogo are friends. Mkhuseli takes Yogo's consumer studies test paper and says: “I have 20 marks more than you do and the sum of both our marks is equal to 112. What are our marks?”

Answer:

Mkhuseli got marks and Yogo got marks for the consumer studies test paper.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.


STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

m= Mkhuseli's marky= Yogo's mark

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

 more than YogoMkhuseli has 20 marksm=y+20(the total) is 112The sum of the marksm+y=112

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

m+y=112(y+20)+y=1122y=11220y=922=46

This means that Yogo's mark is 46.


STEP: Find Mkhuseli's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Yogo's mark and one of the equations we have to find Mkhuseli's mark. The easiest way to do that is to substitute Yogo's mark back into the first equation:

m=y+20=(46)+20=66

The students achieved these marks: Mkhuseli earned 66 and Yogo earned 46.


Submit your answer as: and

ID is: 3278 Seed is: 4292

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 36.

Determine the value of the larger number.

Answer: The larger number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 36. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 36". Remember that sum means addition. So:

n1+n2=36

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

We can solve this using substitution. However, remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)+n2=362n22=362n2=38n2=19

The result is n2=19. Notice that this means that the other number, n1, must be 17, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 36, and 19+17=36.

The larger number is 19.


Submit your answer as:

ID is: 3278 Seed is: 6574

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 40.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 40. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 40". Remember that sum means addition. So:

n1+n2=40

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=40, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=402n1+2=402n1=38n1=19

The result is n1=19. Notice that this means that the other number, n2, must be 21, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 40, and 19+21=40.

The smaller number is 19.


Submit your answer as:

ID is: 3278 Seed is: 8603

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The sum of the numbers is 16.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. For example, you can use n1 and n2. Then write equations using these variables based on the facts given in the question.
STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of 16. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the sum of the numbers is 16". Remember that sum means addition. So:

n1+n2=16

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

We can do this using substitution. If we substitute n2=n1+2 into the equation n1+n2=16, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1+(n1+2)=162n1+2=162n1=14n1=7

The result is n1=7. Notice that this means that the other number, n2, must be 9, because n2=n1+2. This is perfect, because we know that the sum of the numbers is 16, and 7+9=16.

The smaller number is 7.


Submit your answer as:

ID is: 3286 Seed is: 9453

Word problems: checking answers

Suleiman is 11 years younger than his sister, Akeju. In 9 years, Akeju will be 2 times as old as Suleiman. How old is Suleiman now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Suleiman is years old.

numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Suleiman's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Suleiman's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Akeju and Suleiman. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aA=Akeju's ageaS=Suleiman's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Suleiman is 11 years younger than his sister, Akeju." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aS=aA11

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 9 years, Akeju will be 2 times as old as Suleiman." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aA+9=Akeju's age in 9 yearsaS+9=Suleiman's age in 9 years

These are the ages at which Akeju will be 2 times as old as Suleiman. We can put all this together as follows:

in 9 yearsAkeju's age=2×in 9 yearsSuleiman's ageaA+9=2(aS+9)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aA and aS. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aS=aA11. We can substitute this into the second equation and solve for aA.

aA+9=2(aS+9)aA+9=2((aA11)+9)aA+9=2aA413=aA

Terrific: this means that Akeju is 13 years old. But the question asked for us to find Suleiman's age. We can find it using the first equation, which relates the two ages:

aS=aA11=(13)11=2

So we finally got the answer to the question. Suleiman is 2 years old.

The correct answer is: 2.


Submit your answer as:

ID is: 3286 Seed is: 8220

Word problems: checking answers

Matthew is 7 years older than his brother, Richard. In 6 years, Matthew will be 8 times as old as Richard. How old is Richard now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Richard is years old.

one-of
type(string.nocase)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Richard's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Richard's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Matthew and Richard. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aM=Matthew's ageaR=Richard's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Matthew is 7 years older than his brother, Richard." We need to translate that into mathematics. The key word is older, which tells us to use addition to relate the ages.

aM=aR+7

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 6 years, Matthew will be 8 times as old as Richard." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aM+6=Matthew's age in 6 yearsaR+6=Richard's age in 6 years

These are the ages at which Matthew will be 8 times as old as Richard. We can put all this together as follows:

in 6 yearsMatthew's age=8×in 6 yearsRichard's ageaM+6=8(aR+6)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aM and aR. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aM=aR+7. We can substitute this into the second equation and solve for aR.

aM+6=8(aR+6)(aR+7)+6=8(aR+6)aR+13=8aR+4835=7aR5=aR

Terrific: we have the answer.

But wait a minute: aR represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.


Submit your answer as:

ID is: 3286 Seed is: 5705

Word problems: checking answers

Neziswa is 7 years younger than her brother, Fezekile. In 8 years, Fezekile will be 2 times as old as Neziswa. How old is Neziswa now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.
Answer:

Neziswa is years old.

one-of
type(string.nocase)
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Neziswa's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.


STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Neziswa's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Fezekile and Neziswa. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

aF=Fezekile's ageaN=Neziswa's age

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Neziswa is 7 years younger than her brother, Fezekile." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

aN=aF7

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In 8 years, Fezekile will be 2 times as old as Neziswa." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

aF+8=Fezekile's age in 8 yearsaN+8=Neziswa's age in 8 years

These are the ages at which Fezekile will be 2 times as old as Neziswa. We can put all this together as follows:

in 8 yearsFezekile's age=2×in 8 yearsNeziswa's ageaF+8=2(aN+8)

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables aF and aN. The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is aN=aF7. We can substitute this into the second equation and solve for aF.

aF+8=2(aN+8)aF+8=2((aF7)+8)aF+8=2aF+26=aF

Terrific: this means that Fezekile is 6 years old. But the question asked for us to find Neziswa's age. We can find it using the first equation, which relates the two ages:

aN=aF7=(6)7=1

So we finally got the answer to the question. But wait a minute: aN represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.


Submit your answer as:

ID is: 3279 Seed is: 7379

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 35.

Find the value of the larger number.

Answer: The larger number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the larger number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 35. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

n1=the smaller numberwe need to findthis is the numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 35". Remember that product means multiplication. So:

n1n2=35

STEP: Solve the equations for n2
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of n2.

Remember that we want the value of n2 (the larger number). We can start by rearranging the first equation to make n1 the subject. Then the substitution step will remove n1 from the second equation and we can solve for n2. (This is not required - it just makes the solution faster.)

n2=n1+2n22=n1

Now substitute n22 into the other equation and solve for n2.

(n22)n2=35n222n2=35n222n235=0(n27)(n2+5)=0
n2=7 and n2=5

This solution led to two answers for n2. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n2=7.

This means the other number, n1, must be 5, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 35, and 75=35.

The larger number is 7.


Submit your answer as:

ID is: 3279 Seed is: 7832

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 35.

What is the value of the smaller number?

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 35. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 35". Remember that product means multiplication. So:

n1n2=35

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=35, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=35n12+2n1=35n12+2n135=0(n1+7)(n15)=0
n1=7andn1=5

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=5.

This means the other number, n2, must be 7, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 35, and 57=35.

The smaller number is 5.


Submit your answer as:

ID is: 3279 Seed is: 6935

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

  • The numbers are consecutive odd integers.
  • The product of the numbers is 99.

Find the value of the smaller number.

Answer: The smaller number is .
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.
STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of 99. We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

we need to findthis is the numbern1=the smaller numbern2=the larger number

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, 11 and 13 are consecutive odd numbers. Since we defined n1 as the smaller number, n2 must be 2 more than n1.

n2=n1+2

The "+2" skips the even integer which sits between n1 and n2.

The second fact about the numbers tells us that "the product of the numbers is 99". Remember that product means multiplication. So:

n1n2=99

STEP: Solve the equations for n1
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of n1.

If we subsitute n2=n1+2 into the equation n1n2=99, the n2 terms will disappear, leaving n1. Then we can solve for n1, which is exactly what we want.

n1(n1+2)=99n12+2n1=99n12+2n199=0(n1+11)(n19)=0
n1=11andn1=9

This solution led to two answers for n1. However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves n1=9.

This means the other number, n2, must be 11, because n2=n1+2. This is perfect, because we also know that the product of the numbers is 99, and 911=99.

The smaller number is 9.


Submit your answer as:

ID is: 380 Seed is: 8276

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are positive
  • the product of the numbers is 156

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 156.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 156. That means if we multiply the numbers we get 156.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 156. In other words, if we multiply the numbers, we get 156. As an equation this is:

n1n2=156

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=156n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=156(n1)2+n1156=0(n1+13)(n112)=0
n1=13andn1=12

The solutions to the equation are n1=13 or n1=12. This means the first number is either −13 or 12.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −13 and 12 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the n1=13, so the first number is n1=12.

Based on that we can find the second number. The second number is n1+1, which is equal to (12)+1=13.

The two consecutive integers are 12 and 13.


Submit your answer as: and

ID is: 380 Seed is: 542

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are negative
  • the product of the numbers is 132

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 132.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 132. That means if we multiply the numbers we get 132.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 132. In other words, if we multiply the numbers, we get 132. As an equation this is:

n1n2=132

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=132n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=132(n1)2+n1132=0(n1+12)(n111)=0
n1=12andn1=11

The solutions to the equation are n1=12 or n1=11. This means the first number is either −12 or 11.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −12 and 11 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the n1=11, so the first number is n1=12.

Based on that we can find the second number. The second number is n1+1, which is equal to (12)+1=11.

The two consecutive integers are −12 and −11.


Submit your answer as: and

ID is: 380 Seed is: 4534

Word problems: finding consecutive numbers

Here are some facts about two numbers:

  • the numbers are consecutive integers
  • both numbers are positive
  • the product of the numbers is 90

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'
INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.
Answer: The numbers are and .
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by n and n+1. Write an equation based on the fact that the product of those two numbers is 90.


STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of 90. That means if we multiply the numbers we get 90.

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, n is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

n1=the first numbern2=the next number

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is 90. In other words, if we multiply the numbers, we get 90. As an equation this is:

n1n2=90

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if n1 represents the first number, then the next number must be one more than n1.

If:n2=the number following n1then:n2=n1+1

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

n1n2=90n2=n1+1

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for n1.

n1(n1+1)=90(n1)2+n190=0(n1+10)(n19)=0
n1=10andn1=9

The solutions to the equation are n1=10 or n1=9. This means the first number is either −10 or 9.


STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers −10 and 9 are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the n1=10, so the first number is n1=9.

Based on that we can find the second number. The second number is n1+1, which is equal to (9)+1=10.

The two consecutive integers are 9 and 10.


Submit your answer as: and

ID is: 385 Seed is: 400

Word problems: an age-old question

Bonani has a son, Abdulai. Here are some facts about how old Bonani and Abdulai are:

  • Bonani is 5 times as old as Abdulai right now.
  • 8 years from now, Bonani will be 3 times as old as Abdulai.

How old are Bonani and Abdulai now?

Answer:

Bonani is years old and Abdulai is years old.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Bonani and his son, Abdulai. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let b=Bonani's ageLet a=Abdulai's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Bonani is now 5 times as old as Abdulai." As an equation, this is:

Ages now: b=5a

The second piece of information says that in "8 years... Bonani will be 3 times as old as his son." In 8 years Bonani will be b+8 years old, and similarly Abdulai will be a+8 years old. Then:

Ages in 8 years: b+8=3(a+8)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel b.)

b+8=3(a+8)(5a)+8=3a+242a=16a=8

Great! Now we know that Abdulai is 8 years old.


STEP: Use Abdulai's age to find Bonani's age
[−1 point ⇒ 0 / 6 points left]

We can now find Bonani's age. Substitute Abdulai's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

b=5a=5(8)=40

Write your final answer: Bonani is 40 years old and Abdulai is 8 years old.


Submit your answer as: and

ID is: 385 Seed is: 3138

Word problems: an age-old question

Abayomi has a son, Mark. Here are some facts about how old Abayomi and Mark are:

  • Abayomi is 8 times as old as Mark right now.
  • 10 years from now, Abayomi will be 3 times as old as Mark.

How old are Abayomi and Mark now?

Answer:

Abayomi is years old and Mark is years old.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Abayomi and his son, Mark. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let a=Abayomi's ageLet m=Mark's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Abayomi is now 8 times as old as Mark." As an equation, this is:

Ages now: a=8m

The second piece of information says that in "10 years... Abayomi will be 3 times as old as his son." In 10 years Abayomi will be a+10 years old, and similarly Mark will be m+10 years old. Then:

Ages in 10 years: a+10=3(m+10)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel a.)

a+10=3(m+10)(8m)+10=3m+305m=20m=4

Great! Now we know that Mark is 4 years old.


STEP: Use Mark's age to find Abayomi's age
[−1 point ⇒ 0 / 6 points left]

We can now find Abayomi's age. Substitute Mark's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

a=8m=8(4)=32

Write your final answer: Abayomi is 32 years old and Mark is 4 years old.


Submit your answer as: and

ID is: 385 Seed is: 5575

Word problems: an age-old question

Daniel has a son, Babatunde. Here are some facts about how old Daniel and Babatunde are:

  • Daniel is 9 times as old as Babatunde right now.
  • 5 years from now, Daniel will be 5 times as old as Babatunde.

How old are Daniel and Babatunde now?

Answer:

Daniel is years old and Babatunde is years old.

numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.


STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Daniel and his son, Babatunde. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

Let d=Daniel's ageLet b=Babatunde's age

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Daniel is now 9 times as old as Babatunde." As an equation, this is:

Ages now: d=9b

The second piece of information says that in "5 years... Daniel will be 5 times as old as his son." In 5 years Daniel will be d+5 years old, and similarly Babatunde will be b+5 years old. Then:

Ages in 5 years: d+5=5(b+5)

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel d.)

d+5=5(b+5)(9b)+5=5b+254b=20b=5

Great! Now we know that Babatunde is 5 years old.


STEP: Use Babatunde's age to find Daniel's age
[−1 point ⇒ 0 / 6 points left]

We can now find Daniel's age. Substitute Babatunde's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

d=9b=9(5)=45

Write your final answer: Daniel is 45 years old and Babatunde is 5 years old.


Submit your answer as: and

ID is: 3285 Seed is: 5668

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42.195 kilometres. 1,000 people ran the "One Metre at a Time" marathon in Polokwane. The second-place runner finished 16.58 seconds after the winner. The third runner was 15.76 seconds behind the second runner, and the fourth runner was 28.35 seconds behind the third runner. If the second-place runner had an average speed of 19.74 km/h, what was the winning time for the marathon?

Can the answer to the question above be a non-integer number?

Answer:

Can the answer be non-integer?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be non-integer
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. And non-integer values are acceptable for time values. (In other words, an amount of time does not have to be an integer.)

The correct answer is: Yes, it can be a non-integer.


Submit your answer as:

ID is: 3285 Seed is: 9058

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42.195 kilometres. 1,000 people ran the "Sunshine and Roses are for Sissies" marathon in Port Elizabeth. The second-place runner finished 14.83 seconds after the winner. The third runner was 17.23 seconds behind the second runner, and the fourth runner was 21.34 seconds behind the third runner. If the second-place runner had an average speed of 19.69 km/h, what was the winner's average speed for the marathon?

Can the answer to the question above be a non-integer number?

Answer:

Can the answer be non-integer?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be non-integer
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: what was the winner's average speed for the marathon. That means we need a speed value. And non-integer values are acceptable for speed values. (In other words, an amount of speed does not have to be an integer.)

The correct answer is: Yes, it can be a non-integer.


Submit your answer as:

ID is: 3285 Seed is: 7858

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of 42.195 kilometres. 1,000 people ran the "It Ain't Over Till It's Over" marathon in Cape Town. The second-place runner finished 15.19 seconds after the winner. The third runner was 15.16 seconds behind the second runner, and the fourth runner was 28.74 seconds behind the third runner. If the second-place runner had an average speed of 19.51 km/h, what was the winning time for the marathon?

Can the answer to the question above be a non-integer number?

Answer:

Can the answer be non-integer?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.


STEP: Decide if the answer can be non-integer
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a non-integer number or not.

Focus on what the problem asks: what was the winning time for the marathon. That means we need a time value. And non-integer values are acceptable for time values. (In other words, an amount of time does not have to be an integer.)

The correct answer is: Yes, it can be a non-integer.


Submit your answer as:

ID is: 377 Seed is: 2243

Word problems: rectangle facts

The diagonal of a rectangle is 20 cm more than its width. The length of the same rectangle is 10 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is 20 cm more than its width:

d=w+20

The length of the rectangle is 10 cm more than its width:

l=w+10

STEP: Use the theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+20 in for d and w+10 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+20)2=w2+(w+10)2w2+40w+400=w2+(w2+20w+100)0=w220w300

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w220w3000=(w30)(w+10)w=30or w=10

This means that the width of the rectangle is either 30 cm or 10 cm. But the dimensions of a rectangle cannot be negative, so w=30 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+10=(30)+10=40 cm

The width of the rectangle is 30 cm and the length is 40 cm.


Submit your answer as: and

ID is: 377 Seed is: 8357

Word problems: rectangle facts

The diagonal of a rectangle is 22 cm more than its width. The length of the same rectangle is 11 cm more than its width.

What are the width and the length of the rectangle?

Answer: The width is cm and length is cm.
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is 22 cm more than its width:

d=w+22

The length of the rectangle is 11 cm more than its width:

l=w+11

STEP: Use the theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+22 in for d and w+11 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+22)2=w2+(w+11)2w2+44w+484=w2+(w2+22w+121)0=w222w363

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w222w3630=(w33)(w+11)w=33or w=11

This means that the width of the rectangle is either 33 cm or 11 cm. But the dimensions of a rectangle cannot be negative, so w=33 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+11=(33)+11=44 cm

The width of the rectangle is 33 cm and the length is 44 cm.


Submit your answer as: and

ID is: 377 Seed is: 6493

Word problems: rectangle facts

The diagonal of a rectangle is 30 cm more than its width. The length of the same rectangle is 15 cm more than its width.

Determine the width and length of the rectangle.

Answer: The width is cm and length is cm.
numeric
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.


STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the rectangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: d is the diagonal, w is the width, and l is the length.


STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

The diagonal of the rectangle is 30 cm more than its width:

d=w+30

The length of the rectangle is 15 cm more than its width:

l=w+15

STEP: Use the theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the theorem of Pythagoras. Using our variables from the diagram, we can write:

d2=w2+l2

And now we can substitute in the equations from above. Specifically we can substitute w+30 in for d and w+15 in for l. Then expand the binomials and simplify the equation as much as possible.

d2=w2+l2(w+30)2=w2+(w+15)2w2+60w+900=w2+(w2+30w+225)0=w230w675

STEP: Solve the equation for w
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

0=w230w6750=(w45)(w+15)w=45or w=15

This means that the width of the rectangle is either 45 cm or 15 cm. But the dimensions of a rectangle cannot be negative, so w=45 cm.


STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects l to w.

l=w+15=(45)+15=60 cm

The width of the rectangle is 45 cm and the length is 60 cm.


Submit your answer as: and

ID is: 3281 Seed is: 9359

Setting up simultaneous equations

Last week, Ayodeji and Babatunde had a physics test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 146.
  • Ayodeji's mark is 22 less than Babatunde's mark.

Let a represent Ayodeji's mark and b represent Babatunde's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 146.
Ayodeji's mark is 22 less than Babatunde's mark.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Ayodeji's mark and b for Babatunde's mark. So we can break up the first fact like this:

The sum of the marksis146a+b=146

The first fact is equivalent to this equation: a+b=146.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Ayodeji's markis22 less than Babatunde's marka=b22

This equation, a=b22, means that Ayodeji's mark is less that Babatunde's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 146. a+b=146
Ayodeji's mark is 22 less than Babatunde's mark. a=b22

Submit your answer as: and

ID is: 3281 Seed is: 6675

Setting up simultaneous equations

Last week, Ayobami and Bridget had a maths test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 140.
  • Ayobami's mark is 12 less than Bridget's mark.

Let a represent Ayobami's mark and b represent Bridget's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 140.
Ayobami's mark is 12 less than Bridget's mark.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Ayobami's mark and b for Bridget's mark. So we can break up the first fact like this:

The sum of the marksis140a+b=140

The first fact is equivalent to this equation: a+b=140.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Ayobami's markis12 less than Bridget's marka=b12

This equation, a=b12, means that Ayobami's mark is less that Bridget's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 140. a+b=140
Ayobami's mark is 12 less than Bridget's mark. a=b12

Submit your answer as: and

ID is: 3281 Seed is: 752

Setting up simultaneous equations

Last week, Anuoluwapo and Babangida had a physics test. Now they are comparing their marks and they notice these facts:

  • The sum of the marks is 170.
  • Anuoluwapo's mark is 10 less than Babangida's mark.

Let a represent Anuoluwapo's mark and b represent Babangida's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:
Fact about the test scores Equation
The sum of the marks is 170.
Anuoluwapo's mark is 10 less than Babangida's mark.
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.


STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word Meaning
sum +
product ×
is =
consecutive 1 apart
more than add to
less than subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable a for Anuoluwapo's mark and b for Babangida's mark. So we can break up the first fact like this:

The sum of the marksis170a+b=170

The first fact is equivalent to this equation: a+b=170.


STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

Anuoluwapo's markis10 less than Babangida's marka=b10

This equation, a=b10, means that Anuoluwapo's mark is less that Babangida's mark.

The correct answers are:

Fact about the test scores Equation
The sum of the marks is 170. a+b=170
Anuoluwapo's mark is 10 less than Babangida's mark. a=b10

Submit your answer as: and